2.20.3.4.2 Examples for section 3-1-3 (a) Removal of linear term, subcase (ii)

2.20.3.4.2.1 Example \(y^{\prime }=x+3y+7y^{2}\) Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) shows that

\begin{align*} f_{0} & =x\\ f_{1} & =3\\ f_{2} & =7 \end{align*}

Let

\begin{align*} y & =u\left ( x\right ) -v\left ( x\right ) \\ v & =\frac {f_{1}}{2f_{2}}\\ & =\frac {3}{14}\end{align*}

Then \(y=u-\frac {3}{14}\) and \(y^{\prime }=u^{\prime }\). The ode becomes

\begin{align} y^{\prime } & =x+3y+7y^{2}\nonumber \\ u^{\prime } & =x+3\left ( u-\frac {3}{14}\right ) +7\left ( u-\frac {3}{14}\right ) ^{2}\nonumber \\ & =x+3u-\frac {9}{14}+7\left ( u^{2}+\frac {9}{14^{2}}-\frac {6}{14}u\right ) \nonumber \\ & =x+3u-\frac {9}{14}+7u^{2}+\frac {\left ( 7\right ) \left ( 9\right ) }{14^{2}}-\frac {6}{2}u\nonumber \\ & =x-\frac {9}{14}+7u^{2}+\frac {9}{28}\nonumber \\ & =\left ( x-\frac {9}{28}\right ) +7u^{2} \tag {1}\end{align}

We see that the linear term is removed. In the above \(f_{0}=x-\frac {9}{28},f_{1}=0,f_{2}=7\).

Now we will try to solve (1). It is not one of the special cases we have solved before. We can either try to find a particular solution, or if we can’t, then convert it to second order ode and solve it that way.